Ohm’s Law

Potential difference (voltage) and current are fundamental measurements for electric circuits.

• Multiply the potential difference across a device by the current flowing through it to get the power rating for that device.
• Multiply the power rating by the time for which the device is used to get the energy that the device has transferred.

All of that has been covered previously (here). But potential difference and current are not independent quantities: they are linked by a third quantity, called resistance.

Resistance measures how difficult it is for an electric current to flow through a device. The theory of resistance has been covered previously (here) but there is also a simple equation that you are expected to be able to use to calculate resistance. The equation is known as Ohm’s Law and is named after Georg Ohm, who investigated the behaviour of Alessandro Volta’s (then) newly invented electrochemical cell about 200 years ago.

Ohm found that the current flowing through a conductor is directly proportional to the potential difference (voltage) across the conductor. The constant of proportionality is specific to the material used for the conductor and is the property we call resistance.

Ohm’s Law can therefore be written in various different ways, as shown below;

You would normally have to remember Ohm’s Law but its common form will be included on the Revised Equations Sheet for the summer 2022 GCSE examinations (available here). Therefore, all you will have to do is perform any required unit conversions, rearrange the equation (if necessary) and calculate the answer. You might also be required to recall the name of the unit of resistance (the ohm) and its symbol (Ω). Finally, it’s worth noting that the “x” multiplication symbol is normally omitted, so on the Revised Equations Sheet you will see the format V = I R.

Here’s an example of Ohm’s Law in action: suppose you are asked to find the resistance of a lamp that passes a current of 400 mA when connected to a 6 V battery. First you must convert the milliamps to amps (by dividing by 1000) giving 0.400 A. Then divide the potential difference (6 V) by the current (0.4 A) to get the required answer, which is 15 Ω.

In a separate part of the same question, you could also be asked to calculate the power of the lamp and the amount of energy transferred when the lamp is used for five minutes. The equations required to obtain these answers were stated at the start of this article and will be included on the Revised Equations Sheet that you will be given for the exam.

Pause here and do these calculations yourself.

All being well, you should have got anwers of 2.4 W and 720 J respectively, not having forgotten to convert the five minutes into 300 s.

Footnote: If you are wondering where the strange Ω symbol comes from, it is the Greek letter omega, or “O”. We couldn’t use a Roman (normal) letter O as it could easily be interpreted as the number 0 – and that would be very confusing. So we use the Greek symbol Ω instead.